# What is the trough in destructive interference of EM?

1. Sep 8, 2015

### student34

My textbook says that the trough of the electromagnetic wave cancels out when it meets a crest. Okay, that makes sense, but if the crest is the actual photon/wave of energy, what is this trough that can cancel out the energy? It seems to mean that there are "negative energy photons" in between each "normal" photon.

2. Sep 8, 2015

### blue_leaf77

The crest in the electric/magnetic field is the electric/magnetic field vector with the largest magnitude, it's not the energy. The energy in an electromagnetic wave is proportional to the quantity
$$\int \int |\mathbf{E}(\mathbf{r}_\perp,t)|^2 d\mathbf{r}_\perp dt$$.
where $\mathbf{r}_\perp$ are coordinates in the transverse plane.

Last edited: Sep 8, 2015
3. Sep 8, 2015

### student34

I don't know the mathematical explanations of this, so I am wondering if you would you say that the crest is the photon? If so, the trough seems to being negating the photon for the moment that they meet. Is this right?

4. Sep 8, 2015

### DrewD

No, the crest is not a photon. Be careful mixing classical and quantum physics like that. The crest is the maximum value of the E field in one direction and the trough is the maximum in the exact opposite (or minimum in the first). There is not a photon at the crest and an negative or anit- photon at the trough.

5. Sep 9, 2015

### blue_leaf77

It follows from the definition of intensity as the energy density passing through a transverse plane per unit area per unit time. Intensity is proportional to the magnitude squared of electric field. If you want to get the energy, you then need to do the inverse operation as that to obtain the intensity, namely integrate intensity over the transverse area and then integrate with time.